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What is the remainder when 15²⁰¹⁶ is divided by 11, please?

Added 1 minute later
And it needs to be solved without using any computing devices.

4. If a number x = a * 11 + b (where b is the remainder of the division), then x^2016 = A * 11 + b^2016, where A is some integer expression, a polynomial where the argument is the number 11. In fact, if x = 15, then b = 4. x^2016 = A * 11 + 2^4032 We notice: if the remainder were equal to 1, then 1 to any power would remain 1. For powers of two: 2^10 is divisible by 11 with a remainder of 1. Then 2^4032 = (2^10)^403 * 2^2 = (c * 11 + 1)^403 * 2^2 = (C * 11 + 1) * 2^2 It is clear that the final remainder is four.

Yes, 4. Guess, Ment.

Well, since nothing planned for the vacation actually worked out, with Ment's permission, I’ll at least pose a riddle. I myself solved it a very, very long time ago… Now I don’t even remember either the answer or how I solved it, but back then that guy said that I had solved it correctly, although his solution was much shorter and easier than mine. So here is the riddle, and I will try to find, recall, or re-solve it again. I’m not sure if I can solve it again, but here is the riddle. A hollow hemisphere of mass m lies on a table. Water begins to be poured into it from above through a small hole. To what height H will it fill up?

Hermit, is the radius of the hemisphere known? If so, here it is: mg = integrate (h from 0 to H) ρgh*dS(h) = integrate (α from 0 to arcsin(H/R)) ρgRsin(α)*2π(Rcos(α))*dα = integrate (α from 0 to arcsin(H/R)) ρgR*π*sin(2α)*dα = -0.5πρgRcos(2α) (α from 0 to arcsin(H/R)) = -0.5πρgR (cos(2arcsin(H/R))-1) = -0.5πρgR (1-2(H/R)^2-1) = πρgR(H/R)^2 = πρgH^2/R H = sqrt(mR/πρg)

If it's needed, then prove it. I didn't delve into the solution because the dimension of the expression under the root is (kg*m)/((kg/m³)*(m/s²)) = (kg*m)*(m²*s²)/kg = m³*s². After taking the square root, it will be m^3/2*s, and height is measured only in meters to the first power and without any seconds...

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Now I'll also try to understand the solution.

Added 10 minutes ago
From the very beginning, the pressure at height h will not be ρgh, because at the very bottom it will be the largest, and not zero...

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And, if possible, Ment, please provide more details, because the idea of the solution is absolutely correct, but the answer is not yet.

After extracting the square root, it will be m3/2*c, and the height is measured only in meters to the first power, without any seconds...

Yes, all the comments are correct. Tomorrow I will rewrite these integrals more carefully, because right now I've written some nonsense again...

If no one gets ahead of me, of course.

We're here to degrade, we don't need your integrals.

Down with mental activity!

The first mistake was about the pressure being lower at the bottom. Let's correct it: mg = integrate (h from 0 to H) ρg(H-h)*dS(h) = integrate (h from 0 to H) ρgH*dS(h) - integrate (h from 0 to H) ρgh*dS(h) ------------- (1) The second mistake was forgetting R in one place when expanding the small area element dS, and then losing another R somewhere: integrate (h from 0 to H) ρgh*dS(h) = integrate (α from 0 to arcsin(H/R)) ρgRsin(α)*2π(Rcos(α))*Rdα = integrate (α from 0 to arcsin(H/R)) ρgR*π*sin(2α)*R^2*dα = -0.5πρgR^3cos(2α) (α from 0 to arcsin(H/R)) = -0.5πρgR^3 ( cos(2arcsin(H/R))-1) = -0.5πρgR^3 ( 1-2(H/R)^2-1) = πρgRH^2 -------------- (2) The first integral from the sum in (1): integrate (h from 0 to H) ρgH*dS(h) = integrate (α from 0 to arcsin(H/R)) ρgH*2π(Rcos(α))R*dα = 2πρgHR^2 * integrate (α from 0 to arcsin(H/R)) cos(α)dα = 2πρgHR^2 * sin(α) (α from 0 to arcsin(H/R)) = 2πρgHR^2 * H/R = 2πρgRH^2 (3) Let's combine (1), (2), and (3), and remember to cancel out g from both parts, which I forgot to do last time. m = 2πρRH^2 - πρRH^2 = πρRH^2 H = sqrt(m/πρR) If needed, I won't ask for another attempt, as there's nothing more to do for now.

The dimensions are correct here; I'll take a look at the solution now. I solved it again, but my answer seems to be incorrect...

I figured it out again, but the answer I got doesn't seem quite right...

Does R count as one of them?

Why are you simply arithmetically adding the forces ρg(H-h)*dS(h) with which the water presses on infinitely small strips?

Added after 39 seconds

Does R enter into it?

I won't say. ;)

Down with mental activity!

We came here to degrade; we don't need your integrals.

There you can degrade; leave this place for us...:mad::mad::mad: